Simplify and expand the following expression: $ \dfrac{5}{3z - 3}+ \dfrac{5}{4z + 36}+ \dfrac{3}{z^2 + 8z - 9} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{5}{3z - 3} = \dfrac{5}{3(z - 1)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{5}{4z + 36} = \dfrac{5}{4(z + 9)}$ We can factor the quadratic in the third term: $ \dfrac{3}{z^2 + 8z - 9} = \dfrac{3}{(z - 1)(z + 9)}$ Now we have: $ \dfrac{5}{3(z - 1)}+ \dfrac{5}{4(z + 9)}+ \dfrac{3}{(z - 1)(z + 9)} $ The least common multiple of the denominators is: $ 12(z - 1)(z + 9)$ In order to get the first term over $12(z - 1)(z + 9)$ , multiply by $\dfrac{4(z + 9)}{4(z + 9)}$ $ \dfrac{5}{3(z - 1)} \times \dfrac{4(z + 9)}{4(z + 9)} = \dfrac{20(z + 9)}{12(z - 1)(z + 9)} $ In order to get the second term over $12(z - 1)(z + 9)$ , multiply by $\dfrac{3(z - 1)}{3(z - 1)}$ $ \dfrac{5}{4(z + 9)} \times \dfrac{3(z - 1)}{3(z - 1)} = \dfrac{15(z - 1)}{12(z - 1)(z + 9)} $ In order to get the third term over $12(z - 1)(z + 9)$ , multiply by $\dfrac{12}{12}$ $ \dfrac{3}{(z - 1)(z + 9)} \times \dfrac{12}{12} = \dfrac{36}{12(z - 1)(z + 9)} $ Now we have: $ \dfrac{20(z + 9)}{12(z - 1)(z + 9)} + \dfrac{15(z - 1)}{12(z - 1)(z + 9)} + \dfrac{36}{12(z - 1)(z + 9)} $ $ = \dfrac{ 20(z + 9) + 15(z - 1) + 36} {12(z - 1)(z + 9)} $ Expand: $ = \dfrac{20z + 180 + 15z - 15 + 36}{12z^2 + 96z - 108} $ $ = \dfrac{35z + 201}{12z^2 + 96z - 108}$